Help.. probablility I have the best hand...preflop stuff. logarithm??

[Shroomy]

Say I have QQ against 3 Opponents preflop and im first to act.

How do I figure the percentage chance that I have the best hand?

I know there are 36 better hands (all AA and KK combos) out of 2652 possible hands.

I also know I can figure out what cards have X% chance to be the best hand .. but I just cant figure out the reverse.


Any help would be appreciated .. Ill come wash your car or something [img]/images/graemlins/smile.gif[/img]

[Shroomy]

Let me rephrase this one a bit ...

I pick a number out of a hat that I know has the numbers from 1 to 2652 inside of it.

I pick the number 36

There are three people to pick after me

whats the odds that I have picked the lowest number?

[SamIAm]

Ways to be choose 3 friends and have yours be best is C(2616,3). Total ways to choose 3 friends is C(2651,3).

C(2616,3)/C(2651,3).
-Sam

[BruceZ]

[ QUOTE ]
Say I have QQ against 3 Opponents preflop and im first to act.

How do I figure the percentage chance that I have the best hand?

I know there are 36 better hands (all AA and KK combos) out of 2652 possible hands.

[/ QUOTE ]

There are C(50,2) = 50*49/2 = 1225 remaining hands, and there are only 6 AA and 6 KK. You counted each hand twice, e.g., As2c and 2cAs, and you used 52 cards instead of 50. The probability that a particular opponent has one of these hands is 12/1225. The probability that at least 1 of 3 opponents has one of these hands is approximately 3*12/1225 or 1 in 34, but this over counts the rare times that more than 1 player has it. The exact answer is given in a link at the bottom of this post by the inclusion-exclusion principle as 1 in 34.2.

This is different from your second question about drawing numbers because with cards the probability changes depending on which hands each player is dealt. For example, the first opponent may not get AA or KK, but if he gets any hand with an A or K, it becomes less likely for the remaining players to get AA or KK. Still though, in this particular case, we get close to the same answer to the original question if we take 1 minus the probability that players all draw numbers > 12 out of 1225:

1 - (1213/1225 * 1212/1224 * 1211/1223) =~ 1 in 34.3.

[Shroomy]

Thanks guys!

And especially bruce....

That makes me understand why what I thought would work didnt match my numbers

I assummed I could just take 1-( #Opponents * #Better Starting Hands / Total # Statring hands) but it didnt match the calculations I ran via simulation..

The world is a happy place again
oh yeah.. and my numbers were because for the sims, I looped through all possible combinations for each card, so 36 hands but only 12 distint hands disregarding order .. that doesnt affect the problem, but makes more sense of how I stated it.

[BruceZ]

[ QUOTE ]
I assummed I could just take 1-( #Opponents * #Better Starting Hands / Total # Statring hands)

[/ QUOTE ]

You can do that, as a good approximation, but it's 1 - 3*12/1225.


[ QUOTE ]
my numbers were because for the sims, I looped through all possible combinations for each card, so 36 hands but only 12 distint hands disregarding order

[/ QUOTE ]

Disregarding order there are 12 AA and 12 KK, or 24 total. I'm not sure where you get 36, except that 3 opponents * 12 hands = 36 as above.

[Shroomy]

lol thanks again bruce.. my brain is turning to mush, I was looking at my running count collumn.